Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $r = \dfrac{y + 6}{-5y + 50} \div \dfrac{y^2 + 5y - 6}{-4y + 4} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{y + 6}{-5y + 50} \times \dfrac{-4y + 4}{y^2 + 5y - 6} $ First factor the quadratic. $r = \dfrac{y + 6}{-5y + 50} \times \dfrac{-4y + 4}{(y + 6)(y - 1)} $ Then factor out any other terms. $r = \dfrac{y + 6}{-5(y - 10)} \times \dfrac{-4(y - 1)}{(y + 6)(y - 1)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ (y + 6) \times -4(y - 1) } { -5(y - 10) \times (y + 6)(y - 1) } $ $r = \dfrac{ -4(y + 6)(y - 1)}{ -5(y - 10)(y + 6)(y - 1)} $ Notice that $(y - 1)$ and $(y + 6)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ -4\cancel{(y + 6)}(y - 1)}{ -5(y - 10)\cancel{(y + 6)}(y - 1)} $ We are dividing by $y + 6$ , so $y + 6 \neq 0$ Therefore, $y \neq -6$ $r = \dfrac{ -4\cancel{(y + 6)}\cancel{(y - 1)}}{ -5(y - 10)\cancel{(y + 6)}\cancel{(y - 1)}} $ We are dividing by $y - 1$ , so $y - 1 \neq 0$ Therefore, $y \neq 1$ $r = \dfrac{-4}{-5(y - 10)} $ $r = \dfrac{4}{5(y - 10)} ; \space y \neq -6 ; \space y \neq 1 $